Question

By using properties of definite integrals, evaluate the integrals
${\int }_0^{\frac{\pi }{2}}co{s}^{2}dx.$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}co{s}^{2}xdx\Rightarrow I={\int }_0^{\frac{\pi }{2}}co{s}^2(\frac{\pi }{2}-x)dx\Rightarrow I={\int }_0^{\frac{\pi }{2}}si{n}^{2}xdx$
On adding Eqs. (i) and (ii) , we get
$2I={\int }_0^{\frac{\pi }{2}}(si{n}^{2}x+co{s}^{2}x)dx={\int }_0^{\frac{\pi }{2}}dx[\because si{n}^{2}x+co{s}^{2}x=1]=[x{]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}-0\Rightarrow I=\frac{\pi }{4}$